PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 131. Chemical Separations: The Big Picture Classification and comparison of methods2. Fundamentals of Distribution Separations3. Separation Methods Based on Distributions in Discrete Stages Such as solvent extraction and distillation4. Introduction to Distribution Separations in chromatographic methods. The plate theory, the rate theory; van Deemter's equation.Separation Methods Based on Distributions in Discrete Stages (9/18/13)1. Extraction TheoryA. Definition (1) An extraction is a separation technique based on the distribution of solutions between two mutually insoluble phases.(2) The two phases are usually both liquids, but extractions can also be performed when the phases are gas-liquid, gas-solid, liquid-solid, supercritical fluid-solid. B. Extraction process(1) Extraction are usually based on the following procedure: (a) Solute A in phase 1 is mixed with a known volume of phase 2 and allowed to distribute between the two phases. (b) After equilibration, the faction A that has gone into phase 2 is removed and saved for later use or analysis. Any A remaining in phase 1 is either further extracted or discarded. Phase 1Phase 2Phase 1Phase 2Phase 1Phase 2+(2) The phase that originally contains A is known as that raffinate (phase 1). The phase used to extract it is know as extractant (phase 2).(3) If neither phase used in the extraction is a solid, solute A is said to distribute based on its ability to partition between the two phases.Phase 1Phase 2Phase 1Phase 2If the extractant (phase 3 is a solid, solut A is said to distribute based on adsorption to the extractant’s surface. Phase 1Phase 2Phase 1Phase 2C. Simple Extraction(1) The distribution of a simple solution A between two phases, 1 and 2, is described by: Aphase1 Aphase2 KKD = [A]phase1[A]phase2(2) By convention, the above reaction and equilibrium expression are written so thatPhase 1 is the more polar solvent (usually aqueous) and phase 2 is the more non-polar solvent (usually organic).(3) KD is commonly known as the distribution constant. It is constant for a given solute, set of phase, and temperature under ideal condition.Asolid Aphase 1(4) KD can be estimated based on the solubility of A in the two phase.Asolid Aphase 2Ksp,phase1 = [A]phase11Ksp,phase2 = [A]phase21KD = [A]phase1[A]phase2=Ksp,phase2Ksp,phase1(5) KD can be also be predicted using the Hildebrand solubility parameters for phase 1,Phase 2, and solution A.- ln (KD) = (δ1 – δ2)(δ 1 + δ2 - 2δA)RTVi-(6) For non-ideal (i.e. non-dilute) solution, the degree of distribution of A between phase 1 and 2 is no longer constant, but varies with [A].[A]phase1[A]phase2KD = [A]phase1[A]phase2(7) The graph of [A]phase1 vs. [A]phase2 at a given temperature is known as a distribution isotherm.(8) Since the ration of [A]phase2/[A]phase1 (i.e., KD, or the slope of the distribution isotherm) is not constant at higher temperature. In this region K is called distribution coefficient. (9) The condition used in lower end of the above curve (where KD is constant) are referred to as linear conditions. The conditions at the upper end of the cure (where KD varies with the amount of A) is referred to as non-linear conditions.D. Extraction for solutions with multiple forms(1) Many solutes exist in multiple forms forms in solution Examples: acid and bases (HA and A-), solutes that complex with other compounds (A, AL, AL2), and solute that dimerize (A, A2).(2) Since the concentration of the individual forms of A may not be known, its total or analytical concentration (CA) is used CA = [A]form1 + [A]form2 + [A]form3 + [A]form4 +……Examples: (a) For a compound A that undergoes an acid-base reaction: HA H+ + A-CA = [HA] + [A-](b) For a compound A that undergoes an dimerization: 2A A2CA = [A] + 2[A2](3) The distribution of a multiple-form solute (A) is described by using the ration of its analytical concentrations in phases 1 and 2. This value (Dc) is known as the concentration distribution ratio. Dc = CA, phase 2/CA,phase1 (4) Note that for a solute that exists in only one form in either phase, Dc = KDE. Extraction Theory:(1) Single Extraction a. For the extraction of A in phase 1 with phase 2, the fraction of A remaining in phase 1 after a single extraction (fphase1,1) isfphase1,1=(1 + Dc V2/V1)1Where: V1 and V2 = volume f phases 1 and 2.Dc = concentration distribution ratio for AThe value of V2/V1 is also known as phase ration (β). b. The fraction of A extracted into phase 2 is fphase2,1 = 1- fphase1,1a. A problem in extraction many solutes is that Dc may be small. This results in only small amount of A being extracted.b. One way to increase the amount of A extracted is to perform multiple extractions of phase 1 and combine the amounts of A obtained in phase 2 with each extraction.(2) Multiple extractions:c. For multiple extraction of A in phase 1 with fresh portions of phase 2, the fraction remaining in phase 1 after n extractions with equal volumes of phase 2 is fphase1,n=1 + Dc V2/V11nd. The total fraction of A extracted into phase 2 after n extractions(fphase2,n) is fphase2,n = 1- fphase1,n(3) Multiple extractions:a. To remove A from other solution in a mixture using extraction, the recovery of A as well as its purity should be considered. The purity of A(PurityA) in a given phase may be defined as. PurityA= Moles of ATotal moles of all solutes in the phaseb. The purification yield of A express how relative amount of A in a phase has increased as a results of a separation. This is determined by comparing the purity of A in the final vs. initial sample mixture.Purification YieldA=Purity A,phase2Purity A,phase1c. To get a good separation of A from other solutions requires that their values of Dc (or KD) be significantly different (usually 100-fold differences or greater). The Lager the difference, the better the separation.d. One problem with doing multiple extraction of multi-component mixture isthat the purification yield and purity of A decreases as more extractions are performed even through recovery of A increases. This is due to the increased extraction of other components.Phase 1Phase 2[A] = 0.01 M[B] = 1 MV1=V2=10 mLDcA= 10, DcB=0.1Phase 1Phase 2One-Step Extractionfphase1,1=(1 + Dc V2/V1)1fphase1,A=(1 + Dc,A V2/V1)1= 0.091;fphase1,B=(1